Physics - Wave on String Question with Solution | TestHub

Let the equation of a traveling sine wave be:

$y=A\sin (kx-\omega t)$

The velocity of a particle at position $x$ is given by:

$v=\frac{\partial y}{\partial t}=-A\omega \cos (kx-\omega t)$

The speed of the particle is the magnitude of its velocity:

$|v|=A\omega |\cos (kx-\omega t)|$

For two particles at positions $x_1$ and $x_2$ to always have the same speed, their speeds must be equal for all times $t$:

$|\cos (k{x}_1-\omega t)|=|\cos (k{x}_2-\omega t)|$

Squaring both sides to remove the absolute value, we get:

${\cos }^2(k{x}_1-\omega t)={\cos }^2(k{x}_2-\omega t)$

Using the identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$:

$\cos (2k{x}_1-2\omega t)=\cos (2k{x}_2-2\omega t)$

This implies:

$2k{x}_1-2\omega t=2n\pi \pm (2k{x}_2-2\omega t)$

For this equation to hold true for all values of $t$, the time-dependent terms must cancel out. This is only possible if we take the positive sign:

$2k{x}_1-2\omega t=2n\pi +2k{x}_2-2\omega t$

$2k(x_1-x_2)=2n\pi$

$k\Delta x=n\pi$

Substituting the wave number $k=\frac{2\pi }{\lambda }$:

$\frac{2\pi }{\lambda }\Delta x=n\pi \Rightarrow \Delta x=\frac{n\lambda }{2}$

For the minimum non-zero distance between the two particles, we take $n=1$:

$\Delta x=\frac{\lambda }{2}$

Answer: $\frac{\lambda }{2}$